Division of Polynomials. 46. EXAMPLES. 1. Divide a1+ a2 —a—2—a—4 by a2-a-2. Ans. a2+1+a-2. 2. Divide 4 a4 b6+12 a3 b−5+9a2b-4-b-2+2a-2 -a-4b2 by 2 a2b-3+3 ab-2-b-1+a-2b. Ans. 2 a2b-3+3ab2+b-1-a-2b. 47. In the course of algebraic investigations, it is often convenient to separate a quantity into its factors. This is done, when one of the factors is known, by dividing by the known factor, and the quotient is the other factor. And when a letter occurs as a factor of all the terms of a quantity, it is a factor of the quantity, and may be taken out as a factor, with an exponent equal to the lowest exponent which it has in any term, and indeed by means of negative exponents any monomial may be taken out as a factor of a quantity. 48. EXAMPLES. 1. Take out 3 a2b as a factor of 15 a5 b2 + 6 a3 b + 9 a2 b2 + 3 a2 b. Ans. 3 a2b (5 a3b+2a+3b+1). 2. Take out am as a factor of 3 am+1+2 am. Ans. am (3a+2). 3. Take out 2a3 65 c as a factor of 6 a6 b7 c2 + 6 a b3 c -2ab+2-a2 c. Ans. 2 a3 b5 c (3 a3 b2c+3 a-2b3-a-2b-4c-1+ a-3b-5c-1-2-1a-1b-5. 4. Take out b as a factor of a"-1b-b”. Ans. b (an—1— bn−1). Difference of two Powers divisible by Difference of their Roots. 49. Theorem. The difference of two integral positive powers of the same degree is divisible by the difference of their roots. an -b" is divisible by a-b. Thus, Demonstration. Divide an. b" by a-b, as in art. 42, proceeding only to the first remainder, as follows. - b, 1st Remainder = an − 1 b — bn = b (an—1 — bn−1). Now, if the factor an-1-bn-1 of this remainder is divisible by ab, the remainder itself is divisible by a and therefore an-b" is also divisible by a- -b; that is, if the proposition is true for any power, as the (n· also holds for the nth, or the next greater. - 1)st, it But from examples, 10, 11, 12, 13 of art. 43, the proposition holds for the 2d, 3d, 4th, and 5th; and therefore it must be true for the 6th, 7th, 8th, &c. powers; that is, for any positive integral power. may be 50. Corollary. The division of an bn by a -b continued for the purpose of showing the form of the quotient, an -bn an - an−1b| an—1+an¬2 b+an−3 b2+&c..... +ab2-2+bn−1 Division of Polynomials. that is, an-bn a b =an-1+an-2b+an-3 b2+ &c.... +ab+bπ-1, so that each term of the quotient is obtained from the preceding term by diminishing the exponent of a by unity and increasing that of b by unity; and the number of terms is equal to the exponent n. 51. Corollary. If b is put equal to a in the preceding quotient, each of its terms becomes equal to an−1, which gives the peculiar result 52. There are sometimes two or more terms in the divisor, or in the dividend, or in both, which contain the same highest power of the letter according to which the terms are arranged. In this case, these terms are to be united in one by taking out their common factor; and the compound terms thus formed are to be used as simple ones. It is more convenient to arrange the terms which contain the same power of the letter in a column under each other, the vertical bar being used as in art. 17; and to arrange the terms in the vertical columns according to the powers of some letter common to them. 53. EXAMPLES. 1. Divide a2 23-b2 x3—4 a b x2-2 a2x+2abx + a2-b2 by a x — b x b2 a2 - b. In this quotient, the coefficient a+b of x2, the coefficient a-b of x and the term - a+b are successively obtained by dividing the coefficient a2 -b2 of x3 in the dividend, the coefficient a22ab+b2 of x2 in the first remainder, and the coefficient · a2+2ab mainder, by the coefficient a - 62 of x in the second re b of x in the divisor. Ans. (a+b) x2+(a—b) x − (a—b). · - 2. Divide (6 6-10) a1 — (7 b2 — 23 b+20) a3 — (3 b3 -22b231b-5) a2 + (4 b3 — 9 b2+5 b −5) a + b2 -2b by (3b-5) a + b2 —2 b. Ans. 2 a3 (3b-4) a2 + (4 b— 1) a + 1. - · 3. Divide a (b2 — 2 c2) a1 + (b1 — c1) a2 + (b® + 2b4c2b2c4) by a2-b2-c2. Ans. — a1 — (2 b2 — c2) a2 — b1 — b2 c2. |